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Solve the Cauchy problem $u_{tt}-c^2u_{xx}=0$ with conditions $$1. \ u(x,0)=g(x)\\2.\ u_t(x,0)=h(x)$$ Where $h(x)=0,\ g(x)=\begin{cases} 0,\ x<0 \\ 1,\ x\ge0 \end{cases}$.

My solution.

The general solution has the form $u(x,t)=F(x-ct)+G(x+ct)$...(*)

Applying 1. condition we have $G(x)=-F(x),\ x<0$ and $G(x)=1-F(x),\ x\ge0.$

Hence (*) is now $u(x,t)=\begin{cases}F(x-ct)-F(x+ct),\ x+ct<0\\F(x-ct)-F(x+ct)+1,\ x+ct\ge0 \end{cases}$

Applying 2. condition we have $0=-2cF'(x),\ x<0$ and $0=-2cF'(x),\ x\ge0.$

This implies $F'(x)=0$, hence $F=c$ is a constant.

Therefore $u(x,t)=\begin{cases}0,\ x<0\\1,\ x\ge0 \end{cases}$

Is this solution correct?

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2 Answers 2

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Using d'Alembert formula we have that solution to pde is $u(x,t)=\begin{cases} 0,\ x\le-ct\\ \frac{1}{2}, \ -ct<x\le ct\\ 1,\ x>ct \end{cases}$

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No. $u(x,t) = F(x-ct) - F(x+ct)$ for $x+ct < 0$, not for $x < 0$.

EDIT: Write $G(x) = H(x) - F(x)$, so $u(x,t) = F(x-ct) - F(x+ct) + H(x+ct)$, where $H$ is the Heaviside function. Note that $u_t(x,0) = - 2 F'(x) + \delta(x)$, where $\delta$ is the Dirac delta "function"

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  • $\begingroup$ Why $x+ct<0$ and not $x-ct<0$? $\endgroup$
    – user486983
    Mar 26, 2018 at 22:38
  • $\begingroup$ $u(x,t) = F(x-ct) + G(x+ct)$. $G(y) = -F(y)$ when $y < 0$, thus $G(x+ct) = -F(x+ct)$ when $x+ct < 0$. $\endgroup$
    – Robert Israel
    Mar 26, 2018 at 23:44
  • $\begingroup$ Now it's clear, I edited my question to correct it. $\endgroup$
    – user486983
    Mar 27, 2018 at 3:06
  • $\begingroup$ Is my answer now completely correct? $\endgroup$
    – user486983
    Mar 27, 2018 at 3:07
  • $\begingroup$ No, still wrong. $\endgroup$
    – Robert Israel
    Mar 27, 2018 at 5:07

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